Rationalize a surd $\frac{1}{1+\sqrt{2}-\sqrt{3}}$

Question

How can I rationalize the following surd $$\frac{1}{1+\sqrt{2}-\sqrt{3}}$$ What would be the conjugate of the denominator

Answer 1

Rationalise twice, because after rationalising once , there would still remain a surd in the denominator.

First multiply and divide by $1+\sqrt{2}+ \sqrt{3}$

$$\frac{1}{1+\sqrt{2}-\sqrt{3}}\times\frac{1+\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$$ $$= \frac{1+\sqrt{2}+\sqrt{3}}{1+2\sqrt{2}+2-3}=\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$$

Now multiply and divide by $\sqrt{2}$ $$=\frac{\sqrt{2}+2+\sqrt{6}}{4}$$

Answer 2

Assume $(1+√2)=a,√3=b$ and then rationalize it . i assumed so as denominator is simpliefied as $2.√2$

Answer 3

Use twice the conjugate:

$$\begin{aligned} \frac{1}{1+\sqrt{2}-\sqrt{3}} &= \frac{1}{(1+\sqrt{2})-\sqrt{3}}\frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2})+\sqrt{3}}\\ &=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}}\\ &= \frac{2+\sqrt{2}+\sqrt{6}}{4} \end{aligned}$$